First slide

Evaluation of definite integrals

Question

The integral $\int_{0}^{1 / 2} \frac{e^{x} (2 - x^{2})}{(1 - x)^{3 / 2} (1 + x)^{1 / 2}} d x$ is equal to

Moderate

A

$\sqrt{3 e}$
B

$\sqrt{3 e} - 1$
C

$\sqrt{\frac{e}{3}}$
D

$\sqrt{\frac{e}{3}} - 1$

Solution

Let $I = \int_{0}^{1 / 2} \frac{e^{x} (2 - x^{2})}{(1 - x)^{3 / 2} (1 + x)^{1 / 2}} d x$
$\begin{array}{l} = \int_{0}^{1 / 2} e^{x} [\frac{(1 - x^{2}) + 1}{(1 - x) \sqrt{1 - x^{2}}}] d x \\ = \int_{0}^{1 / 2} e^{x} [\frac{\sqrt{1 - x^{2}}}{1 - x} + \frac{1}{(1 - x) \sqrt{1 - x^{2}}}] d x \end{array}$
As $\begin{array}{l} \frac{d}{d x} (\frac{\sqrt{1 - x^{2}}}{1 - x}) = \frac{(1 - x) (\frac{- x}{\sqrt{1 - x^{2}}}) + \sqrt{1 - x^{2}}}{(1 - x)^{2}} \\ = \frac{- x + x^{2} + 1 - x^{2}}{(1 - x)^{2} \sqrt{1 - x^{2}}} = \frac{1}{(1 - x) \sqrt{1 - x^{2}}} \\ ∴ I {= e^{x} (\frac{\sqrt{1 - x^{2}}}{1 - x})]}_{0}^{1 / 2} \\ = \sqrt{e} \frac{\sqrt{3 / 4}}{2 / 2} - 1 = \sqrt{3 e} - 1 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App